> I am not sure why protocol extension need to differ so much and present
dispatch rules that are potentially very confusing.

I think that both the Java and Swift designers want to get the benefits
of multiple inherences, in the meaning time, they don't want to support
multiple objects inherence.

For Java only, as Java allows implicit override, the `@override` is
optional. Related bugs are hard to find if you don't use  `@override`.

For me, the current rule is simple.
1. If there is a `override`, it is "class dynamic binding". If there is no
`override`, it is not. The dynamic type of `self` is irrelevant.
2. The method in 'protocol extension` is default, which means fail safe. It
is not a method of a class until the class implements it.

Zhaoxin

On Wed, Sep 21, 2016 at 4:10 AM, Goffredo Marocchi <pana...@gmail.com>
wrote:

> This would make for hilarious debugging session if you have the misfortune
> of using a third party binary framework that gets updated and provide
> default implementations for some methods in the protocol extensions.
>
> When Java 8 also decided to blur the lines between classes
> (implementations) and interfaces (equivalent of protocols on this side of
> the fence more or less... anyways they represent behaviours/API contracts),
> they also provided a very simple rule to determine on the code would behave
> at runtime.
>
> Extending Interfaces That Contain Default Methods
>
> When you extend an interface that contains a default method, you can do
> the following:
>
>    - Not mention the default method at all, which lets your extended
>    interface inherit the default method.
>    - Redeclare the default method, which makes it abstract.
>    - Redefine the default method, which overrides it.
>
> https://docs.oracle.com/javase/tutorial/java/IandI/defaultmethods.html
>
> I am not sure why protocol extension need to differ so much and present
> dispatch rules that are potentially very confusing.
>
> Casting at runtime should *never* change what implementation of a method
> gets called.
> For a safe by default language the fact that this can occur is quite
> puzzling and worrying. I am not sure the cost of implementing the current
> static/dynamic dispatch rules for default methods in protocol extensions is
> worth it to say it bluntly, but please enlighten me if I am missing an
> obvious huge pink elephant in the room here.
>
> Sent from my iPhone
>
> On 20 Sep 2016, at 16:18, Nevin Brackett-Rozinsky via swift-evolution <
> swift-evolution@swift.org> wrote:
>
> I think there is a deeper issue that may be worth exploring here.
>
> Notably, when one class presents a member function, its subclasses ought
> to use “override” when they reimplement that method themselves, regardless
> of where the superclass’s version comes from.
>
> In the original post, the class “A” expresses (by conforming to protocol
> “Foo”) that it has a member function “bar()”, and “B” is a subclass of “A”
> which wants its own definition of “bar()”.
>
> It seems to me that “B” should not care whether “A” rolled its own
> implementation of “bar()” or used the default implementation provided by
> “Foo”.
>
> From the perspective of “B”, its superclass “A” promises to have a member
> function “bar()”, so “B” should need to use the `override` keyword just
> like it would when overriding any other method.
>
> To illustrate this more clearly, suppose that “Foo” and “A: Foo” are
> defined in a 3rd-party library, while “B: A” is written in a client module.
>
> If the library changes to give “A” its own custom implementation of
> “bar()”, that *should not* affect client code—because the class “A” still
> conforms to “Foo” so it still is known to have a “bar()” method—but right
> now it *does*:
>
> With the status quo, the simple change of moving a function between a
> protocol extension and a conforming class currently requires downstream
> source-code modifications in clients (in this case, adding `override` to
> “B.bar()”).
>
> I propose that `override` should be required in subclasses on any method
> which the superclass proffers as a customization point, no matter the
> provenance of that claim.
>
> Nevin
>
>
>
> On Tue, Sep 20, 2016 at 8:01 AM, Zhao Xin via swift-evolution <
> swift-evolution@swift.org> wrote:
>
>> Thank you to  Игорь Никитин and Adrian Zubarev. Now I can convince myself
>> for this behavior.
>>
>> As Adrian's example shows, there is no `bar()` implemented in `class A`,
>> so there is no `override func bar()` in class B`. My thought that  
>> `self.bar()`
>> would do the same as `(self as! B).bar()`, is called "dynamic binding",
>> which is basing on the `override`. Since there is no `override`, there is
>> no "dynamic binding". I thought "dynamic binding" was basing on dynamic
>> type of `self`. It was not. I was wrong.
>>
>> The behavior is clear now. In class A's `self.bar()`, the runtime finds
>> that there is no implementation of `bar()` in `class A`, so it calls the
>> `bar` in protocol extension. In class A's `(self as! B).bar()`, as `class
>> B` contains the implementation of `bar()`, the runtime calls it.
>>
>> Zhaoxin
>>
>> On Tue, Sep 20, 2016 at 4:21 PM, Adrian Zubarev via swift-evolution <
>> swift-evolution@swift.org> wrote:
>>
>>> I can’t tell you the reason, but to me it feels like it’s doing the
>>> following thing:
>>>
>>> + - - (Type: B) - - +
>>> |                   |
>>> | func bar()    + - + (Type: A) - - +                        < - - - - - - 
>>> - - - - - - - - - - - - - - +
>>> |               |                   |                                       
>>>                            |
>>> + - - - - - - - + func output() + - + (Protocol: Foo) - - +  — - self.bar() 
>>> - + - (self as! B).bar() - +
>>>                 |               |                         |                 
>>>   |
>>>                 + - - - - - - - + (default) func bar()    |  <- - - - - - - 
>>> - +
>>>                                 |                         |
>>>                                 + - - - - - - - - - - - - +
>>>
>>> class A:Foo {
>>>
>>>     func bar() {}
>>>
>>>     func output() {
>>>         print(type(of:self))
>>>         self.bar()
>>>         (self as! B).bar()
>>>     }
>>> }
>>>
>>> class B:A {
>>>
>>>     override func bar() {
>>>         print("I am B.")
>>>     }
>>> }
>>>
>>> Would solve this temporarily.
>>>
>>> And there we are again with the same discussion if custom implementation
>>> of protocol members, which have default implementation, should have the
>>> override keyword or not.
>>>
>>> Imagine your code like this (not valid code):
>>>
>>> protocol Foo {
>>>     func bar()
>>> }
>>>
>>> extension Foo {
>>>     func bar() {
>>>         print("I am bar.")
>>>     }
>>> }
>>>
>>> class A : Foo {
>>>
>>>     func output() {
>>>
>>>         print(type(of:self))
>>>         default.bar() // fallback an call directly the default 
>>> implementation whenever needed
>>>         self.bar() // will print "I am bar." on A().output() but should 
>>> print "I am B." if Self == B
>>>         (self as! B).bar()
>>>     }
>>> }
>>>
>>> class B : A {
>>>
>>>     override func bar() {
>>>         print("I am B.")
>>>     }
>>> }
>>>
>>> I still think default implementations should be called through something
>>> like default. + whenever you override a default implementation you’d
>>> need override. There is a discussion going on: Mark protocol methods
>>> with their protocol. I clearly did not solved your issue, but I might
>>> have wake your interest to participate. ;)
>>>
>>>
>>>
>>> --
>>> Adrian Zubarev
>>> Sent with Airmail
>>>
>>> Am 20. September 2016 um 04:13:22, Zhao Xin via swift-users (
>>> swift-us...@swift.org) schrieb:
>>>
>>> See below code.
>>>
>>> protocol Foo {
>>>
>>>     func bar()
>>>
>>> }
>>>
>>>
>>> extension Foo {
>>>
>>>     func bar() {
>>>
>>>         print("I am bar.")
>>>
>>>     }
>>>
>>> }
>>>
>>>
>>> class A:Foo {
>>>
>>>     func output() {
>>>
>>>         print(type(of:self)) // prints "B".
>>>
>>>         self.bar() // prints "I am bar."
>>>
>>>         (self as! B).bar() // prints "I am B."
>>>
>>>     }
>>>
>>> }
>>>
>>>
>>> class B:A {
>>>
>>>     func bar() {
>>>
>>>         print("I am B.")
>>>
>>>     }
>>>
>>> }
>>>
>>>
>>> let b = B()
>>>
>>> b.output()
>>>
>>>
>>> I thought `self.bar()` would do the same as `(self as! B).bar()`. It
>>> didn't. In my opinion,  `type(of:self) is B.type`, so they should be the
>>> same, shouldn't they?
>>>
>>>
>>> Zhaoxin
>>> _______________________________________________
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>>> swift-us...@swift.org
>>> https://lists.swift.org/mailman/listinfo/swift-users
>>>
>>>
>>> _______________________________________________
>>> swift-evolution mailing list
>>> swift-evolution@swift.org
>>> https://lists.swift.org/mailman/listinfo/swift-evolution
>>>
>>>
>>
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>>
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